Posted by madison on Tuesday, December 4, 2007 at 6:23pm.
how do you write the equilibrium expression for: C3H8 + SO2-><- 3CO2 + 4H2O?
chem - DrBob222, Tuesday, December 4, 2007 at 6:34pm
The equilibrium expression is the product of the products divided by the product of the reactants each raised to a power indicated by the coefficient in the balanced chemical equation. HOWEVER, you must have mixed up two equations because there is no way this will work. I might buy it if the S of the SO2 was not there. I would be willing to bet money that you typed an S (for SO2) but you intended to type 5O2. Please clarify.
chem - madison, Tuesday, December 4, 2007 at 6:37pm
i don't know what it is but that's the equation on my hw
chem - madison, Tuesday, December 4, 2007 at 6:38pm
im sorry your right it is a 5 :)
chem - DrBob222, Tuesday, December 4, 2007 at 6:46pm
how do you write the equilibrium expression for:
C3H8 + 5O2-><- 3CO2 + 4H2O
Keq = (CO2)3(H2O)4/(C3H8)(O2)2.
I am assuming, since this is a combustion reaction, that the H2O is present as a gas. If it isn't, sometimes the water doesn't enter into the equilibrium OR it is essentially constant, in which case it is often omitted from the expression.
chem - DrBob222, Tuesday, December 4, 2007 at 6:58pm
Madison--I made a typo, also. The power of O2 should be to the fifth and not squared. Each concentration is raised to the power of the coefficient in the equation. The coefficient for O2 is 5, therefore (O2)^5. Sorry bout that.
chem - madison, Tuesday, December 4, 2007 at 6:47pm
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